\(\int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx\) [7609]

Optimal result

Integrand size = 33, antiderivative size = 14 \[ \int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx=\log \left (-5+e^x-\frac {1}{e x}\right ) \]

[Out]

ln(exp(x)-5-1/x/exp(1))

Rubi [F]

\[ \int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx=\int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx \]

[In]

Int[(1 + E^(1 + x)*x^2)/(-x - 5*E*x^2 + E^(1 + x)*x^2),x]

[Out]

x + Defer[Int][(-1 - 5*E*x + E^(1 + x)*x)^(-1), x] + Defer[Int][1/(x*(-1 - 5*E*x + E^(1 + x)*x)), x] + 5*E*Def
er[Int][x/(-1 - 5*E*x + E^(1 + x)*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {1+x+5 e x^2}{x \left (-1-5 e x+e^{1+x} x\right )}\right ) \, dx \\ & = x+\int \frac {1+x+5 e x^2}{x \left (-1-5 e x+e^{1+x} x\right )} \, dx \\ & = x+\int \left (\frac {1}{-1-5 e x+e^{1+x} x}+\frac {1}{x \left (-1-5 e x+e^{1+x} x\right )}+\frac {5 e x}{-1-5 e x+e^{1+x} x}\right ) \, dx \\ & = x+(5 e) \int \frac {x}{-1-5 e x+e^{1+x} x} \, dx+\int \frac {1}{-1-5 e x+e^{1+x} x} \, dx+\int \frac {1}{x \left (-1-5 e x+e^{1+x} x\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx=-\log (x)+\log \left (1+5 e x-e^{1+x} x\right ) \]

[In]

Integrate[(1 + E^(1 + x)*x^2)/(-x - 5*E*x^2 + E^(1 + x)*x^2),x]

[Out]

-Log[x] + Log[1 + 5*E*x - E^(1 + x)*x]

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29

method	result	size
risch	\(\ln \left (-\frac {5 x -{\mathrm e}^{x} x +{\mathrm e}^{-1}}{x}\right )\)	\(18\)
norman	\(-\ln \left (x \right )+\ln \left (x \,{\mathrm e} \,{\mathrm e}^{x}-5 x \,{\mathrm e}-1\right )\)	\(20\)
parallelrisch	\(-\ln \left (x \right )+\ln \left (\left (x \,{\mathrm e} \,{\mathrm e}^{x}-5 x \,{\mathrm e}-1\right ) {\mathrm e}^{-1}\right )\)	\(25\)

[In]

int((x^2*exp(1)*exp(x)+1)/(x^2*exp(1)*exp(x)-5*x^2*exp(1)-x),x,method=_RETURNVERBOSE)

[Out]

ln(-(5*x-exp(x)*x+exp(-1))/x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx=\log \left (-\frac {5 \, x e - x e^{\left (x + 1\right )} + 1}{x}\right ) \]

[In]

integrate((x^2*exp(1)*exp(x)+1)/(x^2*exp(1)*exp(x)-5*x^2*exp(1)-x),x, algorithm="fricas")

[Out]

log(-(5*x*e - x*e^(x + 1) + 1)/x)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx=\log {\left (e^{x} + \frac {- 5 e x - 1}{e x} \right )} \]

[In]

integrate((x**2*exp(1)*exp(x)+1)/(x**2*exp(1)*exp(x)-5*x**2*exp(1)-x),x)

[Out]

log(exp(x) + (-5*E*x - 1)*exp(-1)/x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.57 \[ \int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx=\log \left (-\frac {{\left (5 \, x e - x e^{\left (x + 1\right )} + 1\right )} e^{\left (-1\right )}}{x}\right ) \]

[In]

integrate((x^2*exp(1)*exp(x)+1)/(x^2*exp(1)*exp(x)-5*x^2*exp(1)-x),x, algorithm="maxima")

[Out]

log(-(5*x*e - x*e^(x + 1) + 1)*e^(-1)/x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx=\log \left (-5 \, x e + x e^{\left (x + 1\right )} - 1\right ) - \log \left (x\right ) \]

[In]

integrate((x^2*exp(1)*exp(x)+1)/(x^2*exp(1)*exp(x)-5*x^2*exp(1)-x),x, algorithm="giac")

[Out]

log(-5*x*e + x*e^(x + 1) - 1) - log(x)

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {1+e^{1+x} x^2}{-x-5 e x^2+e^{1+x} x^2} \, dx=\ln \left (x\,\mathrm {e}\,{\mathrm {e}}^x-5\,x\,\mathrm {e}-1\right )-\ln \left (x\right ) \]

[In]

int(-(x^2*exp(1)*exp(x) + 1)/(x + 5*x^2*exp(1) - x^2*exp(1)*exp(x)),x)

[Out]

log(x*exp(1)*exp(x) - 5*x*exp(1) - 1) - log(x)